Find Eigenvalues of Matrix: Determinant & Trace Help

[andrey21]

Find the determinant and trace of matrix, and use these and that 6 is an eigenvalue to find all eigenvalues of matrix

( 4,-4,-8
-2, 2, 6
0, 0, -1)




Here's my attempt

determinant = 4((2x-1)-(6x0) - (-4)((-2x-1)-(6x0)) +(-8)((-2x0-(2x0))
=-8+8+0
=0

Trace = sum of leading diagonals

= 4+2+-1
=5

Now I don't know how to use these to establish eigenvalues;

 

[micromass]

Hi andrey21!

You must use the following facts:

• The trace is the sum of all the eigenvalues.
• The determinant is the product of all the eigenvalues.

So, there are three eigenvalues: 6, a and b. Furthermore, the sum of these is the trace and the product is the determinent.

 

[andrey21]

Thank you Micromass

So I have:

6+a+b=5
a+b=-1

6.(a).(b) = 0

So either a or b is zero:

so a=-1 b=0 c=c

6+(-1)+0=5

6(-1)(0)=0

Is this correct?

 

[micromass]

Yes!

Your eigenvalues are -1, 0 and 6.

 

[andrey21]

Thanks Micromass, just another question what are the algebraic and geometric multiplicities? I am not sure what these are

 

[micromass]

Thanks Micromass, just another question what are the algebraic and geometric multiplicities? I am not sure what these are

The algebraic multiplicity of an eigenvalue is number of time you find that eigenvalue. For example, if you found the eigenvalues 3,3 and 0, then 3 has algebraic multiplicty 2 and 0 has algebraic multiplicty 1.
In this case, we have -1, 0 and 6. Thus every eigenvalue appears once, thus all eigenvalues have algebraic multiplicty 1.

Determining the geometric multiplicity is in general quite hard. The geometric multiplicity of an eigenvalue is the dimension ofthe corresponding eigenspace.
In this case, it's easy to determine this, since we always have

1\leq~\text{geom. mult.}~\leq~\text{alg. mult.}

Thus in this case, we find easily that all eigenvalues have geometric multiplicity 1.
Note however, if we would have found 3,3 and 0. Then 0 would have geometric multiplicity 1 and 3 would have geometric multiplicty 1 or 2. We need more information to precisely determine the geometric multiplicity of 3.

 

[andrey21]

So from what you said the algebraic multiplicity is the amount of times the eigenvalue appears in the matrix.

 

[micromass]

So from what you said the algebraic multiplicity is the amount of times the eigenvalue appears in the matrix.

Yes! You can also characterize the algebraic multiplicities by using the characteristic polynomial. But you won't need this here...

 

[andrey21]

Thank you micromass, just one final question.

Given the matrix:

( 1,-1,1
1,-1,1
1,-1,1)

Show that (1,1,1)^T is an eigenvector and find its corresponding eigenvalue.

When proving an eigenvector I usually multiply it by the matrix, then establish the eigenvalue from there. However the T is confusing me

 

[micromass]

The T is just to denote that this is a column and not a row. You must have realized this already, since you multiplied the vector by the matrix and thus you treated the vector as a column. This is what the T denotes.

 

[andrey21]

Ok so multiplying the eigenvector with the matrix i obtain:

(1,1,1) so eigenvalue is 1 and algebraic multiplicity is 6??

 

[micromass]

Ok so multiplying the eigenvector with the matrix i obtain:

(1,1,1) so eigenvalue is 1 and algebraic multiplicity is 6??

(1,1,1) is an eigenvector with eigenvalue 1. But how in Earth did you find algebraic multiplicity ? The algebraic multiplicity can never be larger than 3 in a 3x3-matrix...

 

[andrey21]

I thought algebraic multiplicity was the amount of times the eigenvalue appears in the matrix, which in A 1 appears 6 times. I am sure I've confused something.

 

[HallsofIvy]

An eigenvalue of a matrix does not necessarily appear in a matrix at all- and although 0 appeared twice in the original matrix here, it had algebraic multiplicity one.

It is always possible to write a matrix in diagonal form, with the eigenvalues on the main diagonal or in "Jordan Normal form" which has 1s just above main diagonal and eigenvalues on the main diagonal. In those, each eigenvaue will appear on the main diagonal a number of times equal to its algebraic multiplicity. That is what micromass thought you were talking about.

In this case the characteristic polynomial is (if I have done the calculations correctly) lambda^2(1- lambda). 1 and 0 are eigenvalues. 1 has algebraic multiplicity 1 and 0 has algebraic multiplicity 2. (Since the characteristic polynomial of an n by n matrix has degree n and can be factored into n linear factors (over the complex numbers) the algebraic multiplicities add to n.)

It is easy to show that if <x, y, z> is an eigenvector with eigenvalue 1, then x- y+ z= x, x- y+ z= y, and x- y+ z= z. Those reduce to -y+ z= 0, x- 2y+ z= 0, and x-y= 0. The first says z= y and the third x= y. Putting those into the middle equation, y- 2y+ y= 0 for all y. That is, we can write such an eigenvector <y, y, y>= y<1, 1, 1>. That is a 1 dimensional space- it is spanned by <1, 1, 1>- so the eigenvalue 1 has geometric multiplicity 1. Similarly, if <x, y, z> is an eigenvector with eigenvalue 0, then x- y+ z= 0 so that z= y- x. That is we can write such a vector <x, y, y- x>= <x, 0, -x>+ <0, y, y>= x<1, 0, -1>+ y<0, 1, 1>. <1, 0, 1> and <0, 1, 1> span a two dimensional space so the geometric multiplicity of the eigenvalue 0 is 2.

It is, however, possible for an eigenvalue of algebraic multiplicity higher than 1 to have geometric multiplicity less than the algebraic multiplicity. For example, the matrix [1, 0, 0; 0, 0, 1; 0, 0, 0] has characteristic polynomial (1- lambda)(lambda)^2 and so has 0 as an eigenvalue of algebraic multiplicity 2. But if <x, y, z> is an eigenvector with eigenvalue 0, then we must have x= 0, z= 0, but no condition on y. Any eigenvector corresponding to eigenvalue 0 is of the form <0, y, 0> which is a one dimensional subspace so 0 has geometric multiplicity 1, not 2.

 

[andrey21]

Ok what you have said is all new to me so I'm not sure what the diagonal matrix or 'Jordan normal form' matrix would look like